Answer:
a) the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
Explanation:
Given that;
Gravitational acceleration g = 9.81 m/s²
Mass m = 5 kg
Extension of the spring X = 50 mm = 0.05 m
Spring constant k = ?
we know that;
mg = kX
5 × 9.81 = k(0.05)
k = 981 N/m
a)
Given that; Acceleration of the elevator a = 2 m/s² upwards
Extension of the spring in this situation = X1
Force exerted by the spring = F
we know that;
ma = F - mg
ma = kX1 - mg
we substitute
5 × 2 = 981 × X1 - (5 ×9.81 )
X1 = 0.06019 m
X1 = 60.19 mm
Therefore the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
B)
Acceleration of the elevator = a
The spring is relaxed i.e, it is not exerting any force on the box.
Only the weight force of the box is exerted on the box.
ma = mg
a = g
a = 9.81 m/s² downwards.
Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
It's mass and its age.
About how long will a 2-M star live as a main-sequence star.
1 billion years.
<span>A low mass red giant star's energy comes from.</span>
M = mass of the larger fish =5kg
<span>V = velocity of the larger fish =10m/s</span>
<span>m = mass of the smaller fish =2kg</span>
<span>v = velocity of the smaller fish =10m/s
</span>formula=
<span>MV = mv
5kg*10m/s=2kg*10m/s
biggern mass fish has more momentum
hope this helps
</span>
Answer;
C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.
Explanation;
-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.
-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source